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9-9 The Discriminant Warm Up Warm Up Lesson Presentatio n California Standar ds Preview Preview

9-9 The Discriminant Warm Up Warm Up Lesson Presentation Lesson Presentation California Standards California StandardsPreview

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Page 1: 9-9 The Discriminant Warm Up Warm Up Lesson Presentation Lesson Presentation California Standards California StandardsPreview

9-9 The Discriminant

Warm UpWarm Up

Lesson Presentation

California Standards

PreviewPreview

Page 2: 9-9 The Discriminant Warm Up Warm Up Lesson Presentation Lesson Presentation California Standards California StandardsPreview

9-9 The Discriminant

Warm UpUse the Quadratic Formula to solve each equation.

1. x2 – 5x – 6 = 0

2. 2x2 + 2x – 24 = 0

3. x2 + 10x + 25 = 0 0, –5

3, –4

1, –6

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9-9 The Discriminant

California Standards

22.0 Students use the quadratic formula or factoring techniques or both to determine whether the graph of a quadratic function will intersect the x-axis in zero, one, or two points. 23.0 Students apply quadratic equations to physical problems, such as the motion of an object under the force of gravity.

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9-9 The Discriminant

discriminant

Vocabulary

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9-9 The Discriminant

If the quadratic equation is in standard form, its discriminant is b2 – 4ac. Notice that this is the expression under the square root in the Quadratic Formula.

Recall that quadratic equations can have two, one, or no real solutions. You can determine the number of solutions of a quadratic equation by evaluating the discriminant.

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9-9 The Discriminant

Page 7: 9-9 The Discriminant Warm Up Warm Up Lesson Presentation Lesson Presentation California Standards California StandardsPreview

9-9 The Discriminant

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9-9 The DiscriminantAdditional Example 1A: Using the Discriminant

Find the number of solutions of 3x2 – 2x + 2 = 0 by using the discriminant.

a = 3, b = –2, c = 2

b2 – 4ac = (–2)2 – 4(3)(2)

= 4 – 24

= –20

Identify the values of a, b, and c.

Substitute 3, –2, and 2 for a, b, and c.

Simplify.

b2 – 4ac is negative. There are no real solutions.

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9-9 The DiscriminantAdditional Example 1B: Using the Discriminant

Find the number of solutions of 2x2 + 11x + 12 = 0 by using the discriminant.

a = 2, b = 11, c = 12

b2 – 4ac = 112 – 4(2)(12)

= 121 – 96

= 25

Identify the values of a, b, and c.

Substitute 2, 11, and 12 for a, b, and c.

Simplify.

b2 – 4ac is positive. There are two solutions.

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9-9 The DiscriminantAdditional Example 1C: Using the Discriminant

Find the number of solutions of x2 + 8x + 16 = 0 by using the discriminant.

a = 1, b = 8, c = 16

b2 – 4ac = 82 – 4(1)(16)

= 64 – 64

= 0

Identify the values of a, b, and c.

Substitute 1, 8, and 16 for a, b, and c.

Simplify.

b2 – 4ac is zero. There is one solution.

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9-9 The Discriminant

Check It Out! Example 1a

Find the number of solutions of 2x2 – 2x + 3 = 0 using the discriminant.

a = 2, b = –2, c = 3

b2 – 4ac = (–2)2 – 4(2)(3)

= 4 – 24

= –20

Identify the values of a, b, and c.

Substitute 2, –2, and 3 for a, b, and c.

Simplify.

b2 – 4ac is negative. There are no real solutions.

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9-9 The DiscriminantCheck It Out! Example 1b

Find the number of solutions of x2 + 4x + 4 = 0 using the discriminant.

a = 1, b = 4, c = 4

b2 – 4ac = 42 – 4(1)(4)

= 16 – 16

= 0

Identify the values of a, b, and c.

Substitute 1, 4, and 4 for a, b, and c.

Simplify.

b2 – 4ac is zero. There is one real solution.

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9-9 The Discriminant

Recall that the solutions to a quadratic are the same as the x-intercepts of the related function. The discriminant can be used to find the number of x-intercepts.

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9-9 The DiscriminantAdditional Example 2A: Using the Discriminant to Find

the number of x-Intercepts

Find the number of x-intercepts of y = 2x2 – 9x + 5 using the discriminant.

a = 2, b = –9, c = 5 Identify the values of a, b, and c.

b2 – 4ac = (–9)2 – 4(2)(5)

= 81 – 40

= 41Simplify.

b2 – 4ac is positive.

Therefore, the function y = 2x2 – 9x + 5 has two x-intercepts. The graph intercepts the x-axis twice.

Substitute 2,–9, and 5 for a, b, and c.

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9-9 The DiscriminantAdditional Example 2B: Using the Discriminant to find

the number of x-Intercepts

Find the number of x-intercepts of y = 6x2 – 4x + 5 using the discriminant.

a = 6, b = –4, c = 5 Identify the values of a, b, and c.

b2 – 4ac = (–4)2 – 4(6)(5)

= 16 – 120

= –104 Simplify.

b2 – 4ac is negative.

Therefore, the function y = 6x2 – 4x + 5 has no x-intercepts. The graph does not intercept the x-axis.

Substitute 6, –4, and 5 for a, b, and c.

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9-9 The DiscriminantCheck It Out! Example 2a

Find the number of x-intercepts of y = 5x2 + 3x + 1 by using the discriminant.

a = 5, b = 3, c = 1 Identify the values of a, b, and c.

b2 – 4ac = 32 – 4(5)(1)

= 9 – 20

= –11 Simplify.

b2 – 4ac is negative.

Therefore, the function y = 5x2 + 3x + 1 has no x-intercepts. The graph does not intercept the x-axis.

Substitute 5, 3, and 1 for a, b, and c.

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9-9 The DiscriminantCheck It Out! Example 2b

Find the number of x-intercepts of y = x2 – 9x + 4 by using the discriminant.

a = 1, b = –9, c = 4 Identify the values of a, b, and c.

b2 – 4ac = (–9)2 – 4(1)(4)

= 81 – 16

= 65

Simplify.

b2 – 4ac is positive.

Therefore, the function y = x2 – 9x + 4 has two x-intercepts. The graph intercepts the x-axis twice.

Substitute 1, –9, and 4 for a, b, and c.

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9-9 The Discriminant

The ringer on a carnival strength test is 2 feet off the ground and is shot upward with an initial velocity of 30 feet per second. Will it reach a height of 20 feet? Use the discriminant to explain your answer.

Additional Example 3: Physical Science Application

The height h in feet of an object shot straight up with initial velocity v in feet per second is given by h = –16t2 + vt + c, where c is the initial height of the object above the ground.

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9-9 The DiscriminantAdditional Example 3 Continued

h = –16t2 + vt + c

20 = –16t2 + 30t + 2

0 = –16t2 + 30t + (–18)

b2 – 4ac

302 – 4(–16)(–18) = –252

Substitute 20 for h, 30 for v, and 2 for c.

Subtract 20 from both sides.

Evaluate the discriminant.

Substitute –16 for a, 30 for b, and –18 for c.

The discriminant is negative, so there are no real solutions. The ringer will not reach a height of 20 feet.

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9-9 The Discriminant

If the object is shot straight up from the ground, the initial height of the object above the ground equals 0.

Helpful Hint

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9-9 The DiscriminantCheck It Out! Example 4

What if…? Suppose a weight is shot straight up from the ground with an initial velocity of 20 feet per second. Will it reach the height of 45 feet? Use the discriminant to explain your answer.

h = –16t2 + vt + c

45 = –16t2 + 20t Substitute 45 for h, and 20 for v.

0 = –16t2 + 20t + (–45) Subtract 45 from both sides.

b2 – 4ac

202 – 4(–16)(–45) = –2080

Evaluate the discriminant.

Substitute –16 for a, 20 for b, and –45 for c.

No; for the equation 45 = –16t2 + 20t, the discriminant is negative, so the weight will not ring the bell.

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9-9 The Discriminant

1. Find the number of solutions of 5x2 – 19x – 8 = 0 by using the discriminant.

2. Find the number of x-intercepts of y = –3x2 + 2x – 4 by using the discriminant.

3. An object is shot up from 4 ft off the ground with an initial velocity of 48 ft/s. Will it reach a height of 40 ft? Use the discriminant to explain your answer.

Lesson Quiz

2

2

The discriminant is 0. The object will reach its maximum height of 40 ft once.